Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $p = \dfrac{-4r^2 + 32r}{r^2 - 2r - 3} \times \dfrac{-10r + 30}{-5r + 40} $
Solution: First factor the quadratic. $p = \dfrac{-4r^2 + 32r}{(r - 3)(r + 1)} \times \dfrac{-10r + 30}{-5r + 40} $ Then factor out any other terms. $p = \dfrac{-4r(r - 8)}{(r - 3)(r + 1)} \times \dfrac{-10(r - 3)}{-5(r - 8)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ -4r(r - 8) \times -10(r - 3) } { (r - 3)(r + 1) \times -5(r - 8) } $ $p = \dfrac{ 40r(r - 8)(r - 3)}{ -5(r - 3)(r + 1)(r - 8)} $ Notice that $(r - 8)$ and $(r - 3)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ 40r(r - 8)\cancel{(r - 3)}}{ -5\cancel{(r - 3)}(r + 1)(r - 8)} $ We are dividing by $r - 3$ , so $r - 3 \neq 0$ Therefore, $r \neq 3$ $p = \dfrac{ 40r\cancel{(r - 8)}\cancel{(r - 3)}}{ -5\cancel{(r - 3)}(r + 1)\cancel{(r - 8)}} $ We are dividing by $r - 8$ , so $r - 8 \neq 0$ Therefore, $r \neq 8$ $p = \dfrac{40r}{-5(r + 1)} $ $p = \dfrac{-8r}{r + 1} ; \space r \neq 3 ; \space r \neq 8 $